Mister Exam

Other calculators


sin(4x-3)cos(x+5)

Integral of sin(4x-3)cos(x+5) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  1                           
  /                           
 |                            
 |  sin(4*x - 3)*cos(x + 5) dx
 |                            
/                             
0                             
01sin(4x3)cos(x+5)dx\int\limits_{0}^{1} \sin{\left(4 x - 3 \right)} \cos{\left(x + 5 \right)}\, dx
Detail solution
  1. Rewrite the integrand:

    sin(4x3)cos(x+5)=8sin(5)sin4(x)cos(3)cos(x)8sin3(x)cos(3)cos(5)cos2(x)4sin(5)sin2(x)cos(3)cos(x)+8sin(3)sin(5)sin(x)cos4(x)+4sin(x)cos(3)cos(5)cos2(x)8sin(3)sin(5)sin(x)cos2(x)+sin(3)sin(5)sin(x)8sin(3)cos(5)cos5(x)+8sin(3)cos(5)cos3(x)sin(3)cos(5)cos(x)\sin{\left(4 x - 3 \right)} \cos{\left(x + 5 \right)} = 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} - 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} + 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)} + 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)} - \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}

  2. Integrate term-by-term:

    1. The integral of a constant times a function is the constant times the integral of the function:

      8sin(5)sin4(x)cos(3)cos(x)dx=8sin(5)cos(3)sin4(x)cos(x)dx\int 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\, dx = 8 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin(x)u = \sin{\left(x \right)}.

        Then let du=cos(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        u4du\int u^{4}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

          u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        Now substitute uu back in:

        sin5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

      So, the result is: 8sin(5)sin5(x)cos(3)5\frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5}

    1. The integral of a constant times a function is the constant times the integral of the function:

      (8sin3(x)cos(3)cos(5)cos2(x))dx=8cos(3)cos(5)sin3(x)cos2(x)dx\int \left(- 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        sin3(x)cos2(x)=(1cos2(x))sin(x)cos2(x)\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}

      2. There are multiple ways to do this integral.

        Method #1

        1. Let u=cos(x)u = \cos{\left(x \right)}.

          Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute dudu:

          (u4u2)du\int \left(u^{4} - u^{2}\right)\, du

          1. Integrate term-by-term:

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            1. The integral of a constant times a function is the constant times the integral of the function:

              (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: u33- \frac{u^{3}}{3}

            The result is: u55u33\frac{u^{5}}{5} - \frac{u^{3}}{3}

          Now substitute uu back in:

          cos5(x)5cos3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #2

        1. Rewrite the integrand:

          (1cos2(x))sin(x)cos2(x)=sin(x)cos4(x)+sin(x)cos2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            (sin(x)cos4(x))dx=sin(x)cos4(x)dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

            1. Let u=cos(x)u = \cos{\left(x \right)}.

              Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

              u4du\int u^{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                (u4)du=u4du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                  u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                So, the result is: u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              cos5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: cos5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

          1. Let u=cos(x)u = \cos{\left(x \right)}.

            Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

            u2du\int u^{2}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: u33- \frac{u^{3}}{3}

            Now substitute uu back in:

            cos3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

          The result is: cos5(x)5cos3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #3

        1. Rewrite the integrand:

          (1cos2(x))sin(x)cos2(x)=sin(x)cos4(x)+sin(x)cos2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            (sin(x)cos4(x))dx=sin(x)cos4(x)dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

            1. Let u=cos(x)u = \cos{\left(x \right)}.

              Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

              u4du\int u^{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                (u4)du=u4du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                  u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                So, the result is: u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              cos5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: cos5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

          1. Let u=cos(x)u = \cos{\left(x \right)}.

            Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

            u2du\int u^{2}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: u33- \frac{u^{3}}{3}

            Now substitute uu back in:

            cos3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

          The result is: cos5(x)5cos3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: 8(cos5(x)5cos3(x)3)cos(3)cos(5)- 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      (4sin(5)sin2(x)cos(3)cos(x))dx=4sin(5)cos(3)sin2(x)cos(x)dx\int \left(- 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\right)\, dx = - 4 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin(x)u = \sin{\left(x \right)}.

        Then let du=cos(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        u2du\int u^{2}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

          u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        Now substitute uu back in:

        sin3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      So, the result is: 4sin(5)sin3(x)cos(3)3- \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      8sin(3)sin(5)sin(x)cos4(x)dx=8sin(3)sin(5)sin(x)cos4(x)dx\int 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

      1. Let u=cos(x)u = \cos{\left(x \right)}.

        Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

        u4du\int u^{4}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          (u4)du=u4du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          So, the result is: u55- \frac{u^{5}}{5}

        Now substitute uu back in:

        cos5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

      So, the result is: 8sin(3)sin(5)cos5(x)5- \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5}

    1. The integral of a constant times a function is the constant times the integral of the function:

      4sin(x)cos(3)cos(5)cos2(x)dx=4cos(3)cos(5)sin(x)cos2(x)dx\int 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\, dx = 4 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Let u=cos(x)u = \cos{\left(x \right)}.

        Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

        u2du\int u^{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: u33- \frac{u^{3}}{3}

        Now substitute uu back in:

        cos3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: 4cos(3)cos(5)cos3(x)3- \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      (8sin(3)sin(5)sin(x)cos2(x))dx=8sin(3)sin(5)sin(x)cos2(x)dx\int \left(- 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Let u=cos(x)u = \cos{\left(x \right)}.

        Then let du=sin(x)dxdu = - \sin{\left(x \right)} dx and substitute du- du:

        u2du\int u^{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: u33- \frac{u^{3}}{3}

        Now substitute uu back in:

        cos3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: 8sin(3)sin(5)cos3(x)3\frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      sin(3)sin(5)sin(x)dx=sin(3)sin(5)sin(x)dx\int \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)}\, dx = \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)}\, dx

      1. The integral of sine is negative cosine:

        sin(x)dx=cos(x)\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}

      So, the result is: sin(3)sin(5)cos(x)- \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      (8sin(3)cos(5)cos5(x))dx=8sin(3)cos(5)cos5(x)dx\int \left(- 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{5}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos5(x)=(1sin2(x))2cos(x)\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}

      2. Rewrite the integrand:

        (1sin2(x))2cos(x)=sin4(x)cos(x)2sin2(x)cos(x)+cos(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

      3. Integrate term-by-term:

        1. Let u=sin(x)u = \sin{\left(x \right)}.

          Then let du=cos(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

          u4du\int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        1. The integral of a constant times a function is the constant times the integral of the function:

          (2sin2(x)cos(x))dx=2sin2(x)cos(x)dx\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx

          1. Let u=sin(x)u = \sin{\left(x \right)}.

            Then let du=cos(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

            u2du\int u^{2}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            Now substitute uu back in:

            sin3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

          So, the result is: 2sin3(x)3- \frac{2 \sin^{3}{\left(x \right)}}{3}

        1. The integral of cosine is sine:

          cos(x)dx=sin(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

        The result is: sin5(x)52sin3(x)3+sin(x)\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 8(sin5(x)52sin3(x)3+sin(x))sin(3)cos(5)- 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      8sin(3)cos(5)cos3(x)dx=8sin(3)cos(5)cos3(x)dx\int 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{3}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos3(x)=(1sin2(x))cos(x)\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}

      2. Let u=sin(x)u = \sin{\left(x \right)}.

        Then let du=cos(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        (1u2)du\int \left(1 - u^{2}\right)\, du

        1. Integrate term-by-term:

          1. The integral of a constant is the constant times the variable of integration:

            1du=u\int 1\, du = u

          1. The integral of a constant times a function is the constant times the integral of the function:

            (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: u33- \frac{u^{3}}{3}

          The result is: u33+u- \frac{u^{3}}{3} + u

        Now substitute uu back in:

        sin3(x)3+sin(x)- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 8(sin3(x)3+sin(x))sin(3)cos(5)8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      (sin(3)cos(5)cos(x))dx=sin(3)cos(5)cos(x)dx\int \left(- \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}\right)\, dx = - \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos{\left(x \right)}\, dx

      1. The integral of cosine is sine:

        cos(x)dx=sin(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

      So, the result is: sin(3)sin(x)cos(5)- \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)}

    The result is: 8(sin3(x)3+sin(x))sin(3)cos(5)8(cos5(x)5cos3(x)3)cos(3)cos(5)8(sin5(x)52sin3(x)3+sin(x))sin(3)cos(5)+8sin(5)sin5(x)cos(3)54sin(5)sin3(x)cos(3)3sin(3)sin(x)cos(5)8sin(3)sin(5)cos5(x)5+8sin(3)sin(5)cos3(x)34cos(3)cos(5)cos3(x)3sin(3)sin(5)cos(x)8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} + \frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5} - \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3} - \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)} - \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5} + \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}

  3. Now simplify:

    2cos2(2x)cos(x+2)54cos(2x)cos(x2)5+3cos(x+2)5cos(3x8)6+cos(3x2)2- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}

  4. Add the constant of integration:

    2cos2(2x)cos(x+2)54cos(2x)cos(x2)5+3cos(x+2)5cos(3x8)6+cos(3x2)2+constant- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}


The answer is:

2cos2(2x)cos(x+2)54cos(2x)cos(x2)5+3cos(x+2)5cos(3x8)6+cos(3x2)2+constant- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                                                 /     3         5   \                   /       3         5            \                   /     3            \                      5                         3                         3                         3                         5                 
 |                                                                                  |  cos (x)   cos (x)|                   |  2*sin (x)   sin (x)         |                   |  sin (x)         |                 8*cos (x)*sin(3)*sin(5)   4*cos (x)*cos(3)*cos(5)   4*sin (x)*cos(3)*sin(5)   8*cos (x)*sin(3)*sin(5)   8*sin (x)*cos(3)*sin(5)
 | sin(4*x - 3)*cos(x + 5) dx = C - cos(5)*sin(3)*sin(x) - cos(x)*sin(3)*sin(5) - 8*|- ------- + -------|*cos(3)*cos(5) - 8*|- --------- + ------- + sin(x)|*cos(5)*sin(3) + 8*|- ------- + sin(x)|*cos(5)*sin(3) - ----------------------- - ----------------------- - ----------------------- + ----------------------- + -----------------------
 |                                                                                  \     3         5   /                   \      3          5            /                   \     3            /                            5                         3                         3                         3                         5           
/                                                                                                                                                                                                                                                                                                                                                  
cos(5x+2)10cos(3x8)6-{{\cos \left(5\,x+2\right)}\over{10}}-{{\cos \left(3\,x-8\right) }\over{6}}
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
The answer [src]
  4*cos(1)*cos(6)   sin(1)*sin(6)   sin(3)*sin(5)   4*cos(3)*cos(5)
- --------------- - ------------- - ------------- + ---------------
         15               15              15               15      
5cos8+3cos2303cos7+5cos530{{5\,\cos 8+3\,\cos 2}\over{30}}-{{3\,\cos 7+5\,\cos 5}\over{30}}
=
=
  4*cos(1)*cos(6)   sin(1)*sin(6)   sin(3)*sin(5)   4*cos(3)*cos(5)
- --------------- - ------------- - ------------- + ---------------
         15               15              15               15      
4cos(1)cos(6)15+4cos(3)cos(5)15sin(3)sin(5)15sin(1)sin(6)15- \frac{4 \cos{\left(1 \right)} \cos{\left(6 \right)}}{15} + \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)}}{15} - \frac{\sin{\left(3 \right)} \sin{\left(5 \right)}}{15} - \frac{\sin{\left(1 \right)} \sin{\left(6 \right)}}{15}
Numerical answer [src]
-0.188531945634351
-0.188531945634351
The graph
Integral of sin(4x-3)cos(x+5) dx

    Use the examples entering the upper and lower limits of integration.